Physics For Scientists And Engineers: A Strategic Approach, 4th Edition
Authors: Randall D. Knight
ISBN-13: 978-0133942651
See our solution for Question 4CQ from Chapter 25 from Randall Knight's Physics for Scientists and Engineers.
Problem 4CQ
Chapter:
Problem:
Two protons are launched with the same speed from point 1 inside the parallel-plate capacitor of Figure. Points 2 and 3 are the same distance from the negative plate...
Step-by-Step Solution
Step 1
The points 2 and 3 shown in the Figure are at equal distance from the negative plate, thus these points lie on the same equipotent surface. By the property of the equipotent surface, the potential difference between points lying on the same equipotent surface is zero. Hence the electric potential is same at points 2 and 3.\[{V_2} = {V_3}\]
Step 2: (a) Change in Potential energy
Electric potential energy used when moving from point 1 to 2:\[\begin{array}{l}\Delta {U_{1 - 2}} = q\Delta {V_{1 - 2}}\\ = q\left( {{V_2} - {V_1}} \right)\end{array}\]Electric potential energy used when moving from point 1 to 3:\[\begin{array}{l}\Delta {U_{1 - 3}} = q\Delta {V_{1 - 3}}\\ = q\left( {{V_3} - {V_1}} \right)\end{array}\]Substitute ${V_2} = {V_3}$:\[\begin{array}{l}q\left( {{V_3} - {V_1}} \right) = q\left( {{V_2} - {V_1}} \right)\\\Delta {U_{1 - 3}} = \Delta {U_{1 - 2}}\end{array}\]Change in potential energy along the two paths is same.
Step 3: (b)
When proton moves from point 1 to point 3, or point 1 to point 2, The change in potential energy is same. Thus when potential energy is converted into kinetic energy the proton attains same speed in both cases.
ANSWERS
(a) Change in potential energy is same.
(b) Speed of proton will be same
The points 2 and 3 shown in the Figure are at equal distance from the negative plate, thus these points lie on the same equipotent surface. By the property of the equipotent surface, the potential difference between points lying on the same equipotent surface is zero. Hence the electric potential is same at points 2 and 3.\[{V_2} = {V_3}\]
Step 2: (a) Change in Potential energy
Electric potential energy used when moving from point 1 to 2:\[\begin{array}{l}\Delta {U_{1 - 2}} = q\Delta {V_{1 - 2}}\\ = q\left( {{V_2} - {V_1}} \right)\end{array}\]Electric potential energy used when moving from point 1 to 3:\[\begin{array}{l}\Delta {U_{1 - 3}} = q\Delta {V_{1 - 3}}\\ = q\left( {{V_3} - {V_1}} \right)\end{array}\]Substitute ${V_2} = {V_3}$:\[\begin{array}{l}q\left( {{V_3} - {V_1}} \right) = q\left( {{V_2} - {V_1}} \right)\\\Delta {U_{1 - 3}} = \Delta {U_{1 - 2}}\end{array}\]Change in potential energy along the two paths is same.
Step 3: (b)
When proton moves from point 1 to point 3, or point 1 to point 2, The change in potential energy is same. Thus when potential energy is converted into kinetic energy the proton attains same speed in both cases.
ANSWERS
(a) Change in potential energy is same.
(b) Speed of proton will be same